A level chen問題 acid base equilibria

[匿名]

2008-04-11 8:11 pm

calculate the pH of a 0.01M aqueous solution of sodium ethanoate.
the Ka for ethanoic acid is 1.8 * 10^-5.

[Kw for water at 298K = 1.0 * 10^-14]

請列明詳細計算步驟

thx

回答 (1)

Uncle Michael

2008-04-12 4:24 am

✔ 最佳答案

CH3COO-(aq) + H2O(l) ≒ CH3COOH(aq) + OH-(aq)
First of all, calculate the equilibrium constant, Kh, for the above reaction.
Kh = [CH3COOH][OH-]/CH3COO-]
Kh = ([OH-][H+])/([CH3COOH]/[CH3COO-][H+])
Kh = Kw/Ka
Kh = (1 x 10-14)/(1.8 x 10-5)
Kh = 5.56 x 10-10 M

startgeaa CH3COO-(aq) + H2O(l) ≒ CH3COOH(aq) + OH-(aq)
start(M)eaaaa0.01 OO-(aq) H2O(l) a≒ C0OH(aq) + O0
change(M)aaaa-y OO-(aq) +2O(l) ≒ CH+yOH(aq) + O+y
at eqm(M)aa0.01-y O-(aq) +2O(la) ≒ CHyOH(aq)a + Oy

Kh = y2/(0.01 - y) = 5.56 x 10-10 M
Hence y = 2.36 x 10-6
pOH = -log(y) = 5.63
pH = 14 - pOH = 8.37

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